Case 1
- Angular velocity at the start (rad/s)
- 0rad/s
- Angular velocity at the end (rad/s)
- 20rad/s
- Time the change took (s)
- 4s
5.000rad/s²
Open with these valuesAngular Acceleration Calculator
5.000rad/s²
Result: 5.000 rad/s²Angular acceleration is the change in angular velocity divided by the time it took: from rest to 20 rad/s in 4 seconds is 5 rad/s². Both velocities go in as radians per second, not rpm — divide rpm by 60 and multiply by 2π first. A negative answer simply means the spin is slowing down.
Worked examples
Case 2
- Angular velocity at the start (rad/s)
- 30rad/s
- Angular velocity at the end (rad/s)
- 10rad/s
- Time the change took (s)
- 4s
-5.000rad/s²
Open with these valuesCase 3
- Angular velocity at the start (rad/s)
- -5rad/s
- Angular velocity at the end (rad/s)
- 15rad/s
- Time the change took (s)
- 4s
5.000rad/s²
Open with these valuesHow it's calculated
α = (ω₂ − ω₁) ÷ t
- StepConvert both speeds to rad/s: rpm ÷ 60 × 2π, or rev/s × 2π.
- StepEnter the angular velocity at the start and at the end.
- StepEnter how many seconds the change took.
- ResultRead the answer in rad/s²; a minus sign means it is slowing down.
Reference table
| ω₁, ω₂, t | What is happening | Angular acceleration |
|---|---|---|
| 0, 20, 4 | Spinning up from rest | 5 |
| 10, 40, 6 | Already turning, then faster | 5 |
| 5, 5, 2 | Steady spin, nothing changes | 0 |
| 30, 10, 4 | Braking | -5 |
| 12.5, 7.5, 2.5 | Coasting down gently | -2 |
| 0, 100, 8 | A hard launch | 12.5 |
| -5, 15, 4 | Reversing direction | 5 |
Questions
How do I calculate angular acceleration?
Subtract the starting angular velocity from the final one and divide by the time: α = (ω₂ − ω₁) / t. Radians per second and seconds give rad/s². Going from 0 to 20 rad/s in 4 seconds is 5 rad/s².
Can angular acceleration be negative?
Yes. A negative value means the final angular velocity is smaller than the initial one, so the rotation is slowing down. A wheel dropping from 30 to 10 rad/s over 4 seconds gives −5 rad/s².
My figures are in rpm — what do I enter?
Convert first: rad/s = rpm ÷ 60 × 2π, which is the same as rpm × π ÷ 30. So 3000 rpm is 314.159 rad/s. Entering rpm straight into the field overstates the answer by a factor of 9.55.
Why must the time be greater than zero?
The formula divides the change in angular velocity by the time interval, and dividing by zero is undefined. A time of zero would describe an instant jump rather than a rate.
How does this differ from linear acceleration?
Linear acceleration measures how fast a straight-line speed changes, in m/s²; angular acceleration measures how fast a rotation speed changes, in rad/s². For a point at radius r the tangential linear acceleration is the angular acceleration times r.
Sources and last check
Information, not professional advice.